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the base of an isosceles triangle is 8 feet long. if the altitude is 6 feet long and increasing 3 inches per minute, at what rate are the base angles changing?

ans. 1/52 rad / min

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2 answers
  1. tanθ = h/4
    sec^2θ dθ/dt = 1/4 dh/dt
    52 dθ/dt = 1/4 * 3
    dθ/dt = 3/4 * 1/52

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    oobleck
  2. b = base = 8 ft

    h = altitude = 6 ft

    tan θ = h / ( b / 2 )

    tan θ = 2 h / b

    θ = arctan ( 2 h / b )

    θ is measured in radians

    Using the chain rule:

    u = 2 h / b

    dθ / dh = d [ arctan ( u ) ] / du ∙ du / dh =

    1 / ( 1 + u² ) ∙ ( 2 / b ) = 1 / [ 1 + ( 2 h / b )² ] ∙ ( 2 / b ) =

    1 / ( 1 + 4 h² / b² ) ∙ ( 2 / b ) = 1 / ( b² / b² + 4 h² / b² ) ∙ ( 2 / b ) =

    1 / [ ( b² /+ 4 h² ) / b² ] ∙ ( 2 / b ) = b² / ( b² /+ 4 h² ) ∙ ( 2 / b ) =

    b ∙ b / ( b² /+ 4 h² ) ∙ ( 2 / b )

    dθ / dh = 2 b / ( b² + 4 h² )

    In this case:

    dθ / dt = 2 b / ( b² + 4 h² ) = 2 ∙ 8 / ( 8² + 4 ∙ 6² ) =

    16 / ( 64 + 4 ∙ 36 ) = 16 / ( 64 + 144 ) = 16 / 208 = 16 ∙ 1 / 16 ∙ 13 = 1 / 13

    dθ / dh = 1 / 13 rad / ft

    The altitude increasing 3 inches per minute means dh / dt = 3 in / min

    1 ft = 12 in

    1 in = 1 / 12 ft

    3 in = 3 / 12 ft

    3 in = 3 ∙ 1 / 3 ∙ 4 ft

    3 in = 1 / 4 ft

    3 in / min = 1 / 4 ft / min

    The altitude increasing dh / dt = 1 / 4 ft / min.

    dθ / dt =( dθ / dt) ∙ ( dh / dh )

    dθ / dt = ( dθ / dh ) ∙ ( dh / dt )

    dθ / dt = ( 1 / 13 ) rad / ft ∙ ( 1 / 4 ) ft / min

    dθ / dt = ( 1 / 13 ) rad ∙ ( 1 / 4 ) / min

    dθ / dt = 1 / 13 ∙ 4 rad / min

    dθ / dt = 1 / 52 rad / min

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